Question

question 32 (1 point)
the point (-4,-6) is on the terminal arm of dc. which is the set of exact primary
trigonometric ratios for the angle?
a)
sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},
tan c = \frac{3}{2}
b)
sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},
tan c = \frac{3}{2}
c)
sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},
tan c = \frac{3}{2}
d)
sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},
tan c = \frac{3}{2}

Explanation:

Step1: Calculate the hypotenuse \( r \)

For a point \((x,y)=(-4,-6)\) on the terminal arm of an angle, use the formula \( r=\sqrt{x^{2}+y^{2}} \).

$$ r=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13} $$

Step2: Calculate \(\sin C\), \(\cos C\) and \(\tan C\)

The trigonometric ratios are defined as \(\sin C=\frac{y}{r}\), \(\cos C=\frac{x}{r}\) and \(\tan C=\frac{y}{x}\).

• \(\sin C=\frac{-6}{2\sqrt{13}}=\frac{-3}{\sqrt{13}}=\frac{-3\sqrt{13}}{13}\)
• \(\cos C=\frac{-4}{2\sqrt{13}}=\frac{-2}{\sqrt{13}}=\frac{-2\sqrt{13}}{13}\)
• \(\tan C=\frac{-6}{-4}=\frac{3}{2}\)

Answer:

d) \(\sin C =-\frac{3\sqrt{13}}{13},\cos C =-\frac{2\sqrt{13}}{13},\tan C=\frac{3}{2}\)